# CHAPTER19:TURBULENT BOUNDARY LAYERS¶

## Example E01 : Pg 612¶

In :
# All the quantities are expressed in SI units
# (a)
from math import sqrt
Re_c = 1.36e7;                    # as obtained from ex. 18.1a
rho_inf = 1.22;                   # freestream air denstiy
S = 40.;                           # plate planform area
# hence, from eq.(19.2)
Cf = 0.074/Re_c**0.2;
V_inf = 100.;
# hence, for one side of the plate
D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;
# the total drag on both the surfaces is
D = 2.*D_f;
print"The total frictional drag is: (a)D =",D,"N"
# (b)
Re_c = 1.36e8;                    # as obtained from ex. 18.1b
# hence, from fig 19.1 we have
Cf = 1.34*10.**-3.;
V_inf = 1000.;
# hence, for one side of the plate
D_f = 1./2.*rho_inf*V_inf**2.*S*Cf;
# the total drag on both the surfaces is
D = 2.*D_f;
print"(b) D =",D,"N"

The total frictional drag is: (a)D = 1351.89748485 N
(b) D = 65392.0 N


## Example E02 : Pg 612¶

In :
# All the quantities are expressed in SI units
# from ex 18.2
from math import sqrt
Re_c_star = 3.754e7;                # Reynolds number at the trailing edge of the plate
rho_star = 0.574;
ue = 1000.;                          # velocity of the upper plate
S = 40.;                             # plate planform area
# from eq.(19.3) we have
Cf_star = 0.074/Re_c_star**0.2;
# hence, for one side of the plate
D_f = 1./2.*rho_star*ue**2.*S*Cf_star;
# the total drag on both the surfaces is
D = 2.*D_f;
print"The total frictional drag is:D =",D,"N"

The total frictional drag is:D = 51916.421508 N


## Example E03 : Pg 615¶

In :
# All the quantities are expressed in SI units
import math
Me = 2.94;                          # mach number of the flow over the upper plate
ue = 1000.;
Te = 288.;                           # temperature of the upper plate
ue = 1000.;                          # velocity of the upper plate
S = 40.;                             # plate planform area
Pr = 0.71;                          # Prandlt number of air at standard condition
gam = 1.4;                          # ratio of specific heats

# the recovery factor is given as
r = Pr**(1./3.);

# for M = 2.94
T_aw = Te*(1.+r*(2.74-1.));
T_w = T_aw;                          # since the flat plate has an adiabatic wall

T_star = Te*(0.5*(1.+T_w/Te) + 0.16*r*(gam-1.)/2.*Me**2.);

# from the equation of state
p_star=1.
R=1.
rho_star = p_star/R/T_star;

# from eq.(15.3)
mue0=1.
T0=1.
c=1.
mue_star = mue0*(T_star/T0)**1.5*(T0+110.)/(T_star+110.);

# thus
Re_c_star = rho_star*ue*c/mue_star;

# from eq.(18.22)
Cf_star = 0.02667/Re_c_star**0.139;

# hence, the frictional drag on one surface of the plate is
D_f = 1./2.*rho_star*ue**2.*S*Cf_star;

# thus, the total frictional drag is given by
D = 2.*D_f;

print"The total frictional drag is:D =",D,"N"

The total frictional drag is:D = 4967.70450221 N